[2 Mark Question Answer]

Question 12 Marks

From the following figure,

prove that $: AD=CE$.

Answer

In $\triangle ACD$ and $\triangle CBE,$
$DC= CB .......[$Given$]$
$AC = BE .......[$Given$]$
$\angle ACD = \angle CBE .......[$Proved Earlier$]$
$\therefore \triangle ACD ≅ \triangle CBE .......[SAS$ Criterion$]$
$\Rightarrow AD= CE ......[ \text{c.p.c.t.} ]$

View full question & answer→

Question 22 Marks

From the following figure,

prove that $: \angle ACD =\angle CBE$

Answer

In $\triangle ACB,$
$AC = AC ....[$Given$]$
$\therefore \angle ABC = \angle ACB ...(i) [$angles opposite to equal sides are equal$]$
$\angle ACD + \angle ACB = 180^\circ ......(ii) [DCB$ is a straight line$]$
$\angle ABC + \angle CBE = 180^\circ ….(iii)[ ABE$ is a straight line $]$
Equating $(ii)$ and $(iii)$
$\angle ACD + \angle ACB = \angle ABC + \angle CBE$
$\Rightarrow \angle ACD + \angle ACB = \angle ACB + \angle CBE ......[$From $(i)]$
$\Rightarrow \angle ACD = \angle CBE.$

View full question & answer→

Question 32 Marks

In the given figure$; AE$ bisects exterior angle $CAD$ and $AE$ is parallel to $BC.$

Prove that $: A B=A C$.

Answer

Since $AE || BC$ and $DAB$ is the transversal.
$\therefore \angle DAE = \angle ABC = \angle B .......[$Corresponding angles$]$
Since $AE || BC$ and $AC$ are the transversals.
$\therefore \angle CAE = \angle ACB = \angle C ......[$Alternate angles$]$
But $AE$ bisects $\angle CAD,$
$\therefore \angle DAE = \angle CAE$
$\Rightarrow \angle B =\angle C$
$\Rightarrow AB = AC .......[$Sides opposite to equal angles are equal$]$

View full question & answer→

MATHEMATICS [2 Mark Question Answer]

Test yourself on this topic