ABC is a right angle triangle such that AB=AC and bisector of angleC intersects the side AB at D. Prove that AC+AD=BC
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Step-by-step explanation:
It seems AC + AD = √2 AC = BC.
Perhaps there is a mistake in the given problem..
See the Diagram.
Draw a perpendicular from D onto BC to meet it at E.
∠EDC = ∠ADC as ∠E = 90° = ∠A and ∠ACD = ∠ECD as CD is a bisector of ∠C.
ΔADC and ΔEDC are congruent, as CD is a common side.
=> AC = CE and AD = DE
In ΔBDE, ∠BDE = ∠DBE
So BE = DE => BE = AD
AC + AD = CE + EB
= BC
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