In the following diagram; A D=A B and A E bisect A. Prove that:(i) BE = DE,(ii) A B D> C

Const: Join ED. In AOB and AOD, A B=A D ...[ Given] A O=A O ....[ Common ] BAO = DAO ....[ AO is bisector of A] AOB AOD ....[SAS criterion ] Hence, B O=O D ...(i) [ c.p.c.t. ] AOB = AOD ...(ii)[ c.p.c.t. ] ABO = ADO ABD = ADB ...(iii)[ c.p.c.t. ] Now, AOB = DOE ...[Vertically opposite angles] AOD = BOE ...[Vertically opposite angles] BOE = DOE ...(iv)[ From (ii)] (i) In BOE and DOE, B O=C D ...[ From (i) ] OE = OE ...[ Common ] BOE = DOE ... [ From (iv) ][ SAS criterion] Hence, BE = DE ...[ c.p.c.t. ] (ii) In BCD ADB = C+ C B D [ Ext. angle = sum of opp. int. angles ] A D B> C ABD > C ...[ From (iii) ]

In the following diagram; A D=A B and A E bisect A. Prove that:(i…

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Question

In the following diagram; $A D=A B$ and $A E$ bisect $\angle A$.

Prove that:$(i) BE = DE,(ii) \angle A B D>\angle C$

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Answer

Const: Join $ED.$
In $\triangle AOB$ and $\triangle AOD$,
$A B=A D\dots ...[$ Given$]$
$A O=A O\dots ....[$ Common $]$
$\angle BAO =\angle DAO\dots ....[ AO$ is bisector of $A]$
$\therefore \triangle AOB \cong \triangle AOD ....[\text{SAS}$ criterion $]$
Hence,
$B O=O D\dots ...(i) [\text{ c.p.c.t.} ]$
$\angle AOB =\angle AOD \dots...(ii)[ \text{c.p.c.t.} ]$
$\angle ABO =\angle ADO \Rightarrow \angle ABD =\angle ADB \dots...(iii)[ \text{c.p.c.t. }]$
Now,
$\angle AOB =\angle DOE\dots ...[$Vertically opposite angles$]$
$\angle AOD =\angle BOE\dots ...[$Vertically opposite angles$]$
$\angle BOE =\angle DOE\dots ...(iv)[$ From $(ii)]$
$(i)$ In $\triangle BOE$ and $\triangle DOE$,
$B O=C D \dots...[$ From $(i) ]$
$OE = OE \dots...$[ Common $]$
$\angle BOE =\angle DOE\dots ... [$ From $(iv) ][ \text{SAS}$ criterion$]$
Hence, $BE = DE\dots ...[\text{ c.p.c.t.} ]$
$(ii)$ In $BCD$
$\angle ADB =\angle C+\angle C B D$
$\ldots[$ Ext. angle $=$ sum of opp. int. angles $]$
$\Rightarrow \angle A D B>\angle C$
$\Rightarrow \angle ABD >\angle C \dots...[$ From $(iii) ]$