1 Marks Question

Question 11 Mark

$ABC$ is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to sides $AC$ and $AB$ respectively (See figure). Show that these altitudes are equal.

Answer

In $\triangle ABE$ and $\triangle ACF,$
$\angle A= \angle A [$Common$]$
$\angle AEB = \angle AFC = [90^\circ ]$
$AB = AC [$Given$]$
$\therefore \triangle ABE ≅ \triangle ACF [$By $ASA$ congruency$]$
$\therefore BE = CF [$By $C.P.C.T.]$
So Altitudes are equal.

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Question 21 Mark

Line $l$ is the bisector of an angle $\angle A$ and $B$ is any point on $l. BP$ and $BQ$ are perpendicular from $B$ to the arms of $\angle A.$

Show that:
$i. \triangle APB \cong \triangle AQB$
$ii. BP = BQ$ or $B$ is equidistant from the arms of $\angle A.$

Answer

$i$. In $\triangle APB$ and $\triangle AQB$
$ \angle BAP = \angle BAQ ...[$As l bisects $\angle A]$
$ \angle BPA = \angle BQA ...[$Each $90°]$
$AB = AB ...[$Common$]$
$\therefore \triangle APB \cong \triangle AQB$ proved $...[\text{AAS}$ property$] ...(1)$
$ii. \triangle APB \cong \triangle AQB ...[$From $(1)]$
$ \therefore BP = BQ ...[\text{c.p.c.t.}]$

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Question 31 Mark

$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$. Show that $\Delta ABC \cong \Delta CDA.$

Answer

Given : $l || m$ and $p || q$
To prove : $DABC \cong DCDA$
Proof : $l || m$ and $p || q . . . . [$Given$]$
In $DABC$ and $DCDA$
$ \angle BAC = \angle DCA . . . . . [$Alternate interior angles as $AB || DC]$
Similarly, $\angle ACB = \angle CAD . . . [$Alternate interior angles as $BC || DA]$
$AC = DA . . . [$Common$]$
$DABC \cong DCDA [$By $ASA$ congruency$]$

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Question 41 Mark

$AD$ and $BC$ are equal perpendiculars to a line segment $AB.$ Show that $CD$ bisects $AB ($See figure$)$

Answer

In $\triangle BOC$ and $\triangle AOD,$
$\angle OBC = \angle OAD = 90^\circ [$Given$]$
$ \angle BOC = \angle AOD [$Vertically Opposite angles$]$
$BC = AD [$Given$]$
$ \therefore \triangle BOC \cong \triangle AOD [$By $ASA$ congruency$]$
$ \Rightarrow OB = OA$ and $OC = OD [$By $C.P.C.T.]$

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Question 51 Mark

$D$ is a point on side $BC$ of $\triangle ABC$ such that $AD = AC ($see Fig.$).$ Show that $AB > AD.$

Answer

In $\triangle DAC, AD = AC ($Given$)$
So, $\angle ADC = \angle ACD ($Angles opposite to equal sides$)$
Now, $\angle ADC$ is an exterior angle for $\triangle ABD.$
So, $\angle ADC > \angle ABD$
or, $\angle ACD > \angle ABD$
or, $\angle ACB > \angle ABC$
So, $AB > AC ($Side opposite to larger angle in $\triangle ABC)$
or, $AB > AD (AD = AC)$

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Question 61 Mark

In an isosceles triangle $ABC$ with $AB = AC, D$ and $E$ are points on $BC$ such that $BE = CD ($see Fig.$).$ Show that $AD = AE$

Answer

In $\triangle ABD$ and $\triangle ACE,$
$AB = AC ($Given$) ...(1)$
$ \angle B = \angle C ($Angles opposite to equal sides$) ...(2)$
Also, $BE = CD$
So, $BE – DE = CD – DE$
That is, $BD = CE ...(3)$
So, $\triangle ABD \cong \triangle ACE ($Using $(1), (2), (3)$ and $SAS$ rule$).$
This gives $AD = AE (CPCT)$

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Question 71 Mark

$E$ and $F$ are respectively the mid-points of equal sides $AB$ and $AC$ of $\triangle ABC $(see Fig.$)$. Show that $BF = CE$

Answer

In $\triangle ABF$ and $\triangle ACE,$
$AB = AC ($Given$)$
$ \angle A = \angle A ($Common$)$
$AF = AE ($Halves of equal sides$)$
So, $\triangle ABF \cong \triangle ACE (SAS$ rule$)$
Therefore, $BF = CE (CPCT)$

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Question 81 Mark

In $\triangle ABC,$ the bisector $AD$ of $\angle A$ is perpendicular to side $BC ($See figure$).$ Show that $AB = AC$ and $\triangle ABC$ is isosceles.

Answer

In $\triangle ADB$ and $\triangle ADC,$
$BD = CD [AD$ bisects $BC]$
$ \angle ADB = \angle ADC = 90^\circ [AD \bot BC]$
$AD = AD [$Common$]$
$ \therefore \triangle ABD \cong \triangle ACD [$By $SAS$ congruency$]$
$ \Rightarrow AB = AC [$By $C.P.C.T.]$
Therefore, $ABC$ is an isosceles triangle.

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Question 91 Mark

In Fig., line segment $AB$ is parallel to another line segment $CD. O$ is the mid$-$point of $AD.$ Show that:
$i. \triangle AOB \cong \triangle DOC$
$ii. O$ is also the mid$-$point of $BC.$

Answer

$i.$Consider $\triangle AOB$ and $\triangle DOC.$
$ \angle ABO = \angle DCO ($Alternate angles as $AB \| CD$ and $BC$ is the transversal$)$
$ \angle AOB = \angle DOC ($Vertically opposite angles$)$
$OA = OD ($Given$)$
Therefore, $\triangle AOB \cong \triangle DOC (\text{AAS}$ rule$)$
$ii.OB = OC (\text{CPCT})$
So, $O$ is the mid$-$point of $BC$

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