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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.

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Solution

Given: In right angled ΔABC,AB=AC and CD is the bisector of C.
Construction: Draw DE BC.
To prove: AC + AD = BC
Proof: In right angled ΔABC,AB=AC and BC is a hypotenuse [given]
A=90
In ΔDAC and ΔDEC, A=3=90

1=2 [given, CD is the bisector of C]
DC = DC [common sides]
ΔDACΔDEC [by AAS congruence rule]
DA=DE. [by CPCT] ...(i)
and AC = EC ...(ii)
In ΔABC,AB=AC
C=B
[angles opposite to equal sides are equal] ...(iii)
Again, in ΔABC,A+B+C=180
[by angle sum property of a triangle]
90+B+B=180 [from Eq. (iii)]
2B=180902B=90B=45In ΔBED, 5=180(B+4)
[by angle sum property of a triangle]
=180(45+90)=180135=45B=5
DE=BE[ sides opposite to equal angles are equal]...(iv)
From Eqs. (i) and (iv).
DA = DE = BE ...(v)
BC=CE+EB
= CA + DA [from Eqs. (ii) and (v)]
AD+AC=BC

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