Question

Question 18
If ABC is a right angled triangle such that AB = AC and bisector of angle C intersects the side AB at D, then prove that AC + AD = BC.

Answer 1

Given: In right angled $\Delta ABC,AB=AC$ and CD is the bisector of $\angle C.$
Construction: Draw DE $\perp$ BC.
To prove: AC + AD = BC
Proof: In right angled $\Delta ABC,AB=AC$ and BC is a hypotenuse [given]
$\angle A={90}^{\circ }$
In $\Delta DACand\Delta DEC,\angle A=\angle 3={90}^{\circ }$

$\angle 1=\angle 2$ [given, CD is the bisector of $\angle C$]
DC = DC [common sides]
$\therefore \Delta DAC\cong \Delta DEC$ [by AAS congruence rule]
$\Rightarrow DA=DE.$ [by CPCT] ...(i)
and AC = EC ...(ii)
In $\Delta ABC,AB=AC$
$\angle C=\angle B$
[angles opposite to equal sides are equal] ...(iii)
Again, in $\Delta ABC,\angle A+\angle B+\angle C={180}^{\circ }$
[by angle sum property of a triangle]
$\Rightarrow {90}^{\circ }+\angle B+\angle B={180}^{\circ }$ [from Eq. (iii)]
$\Rightarrow 2\angle B={180}^{\circ }-{90}^{\circ }\Rightarrow 2\angle B={90}^{\circ }\Rightarrow \angle B={45}^{\circ }In\Delta BED,\angle 5={180}^{\circ }-(\angle B+\angle 4)$
[by angle sum property of a triangle]
$={180}^{\circ }-({45}^{\circ }+{90}^{\circ })={180}^{\circ }-{135}^{\circ }={45}^{\circ }\therefore \angle B=\angle 5$
$\Rightarrow DE=BE$[$\because$ sides opposite to equal angles are equal]...(iv)
From Eqs. (i) and (iv).
DA = DE = BE ...(v)
$\because BC=CE+EB$
= CA + DA [from Eqs. (ii) and (v)]
$\therefore AD+AC=BC$