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DeltaABC is a right triangle right angle...

`DeltaABC` is a right triangle right angled at A such that `AB = AC` and bisector of `/_C` intersects the side AB at D. Prove that `AC + AD = BC`.

Text Solution

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Given `A DeltaABC` in which `AB = AC, angleA=90^(@) and Cd` bisect `angleC`.
TO PROVE `AC+AD=BC.`
CONSTRUCTION Draw `DE_|_BC`.
PROOF `DeltaDAC~=DeltaDEC` as `DC=DC, angle1 =angle2, angleA=angle3=90^(@)`.
`therefore" "DA=DE and AC=EC.`
Now, `AB=AC rArr angleB = angleC.`
`angleA+angleB+angleC=180^(@) rArr 90^(@) +angleB +angleB=180^(@) rArr angleB=45^(@)`.
In `DeltaBED, angle4+angleB=90^(@) rArr angle4+45^(@)=90^(@) rArr angle4=45^(@)`.
In `DeltaBDE, angle4=angleB rArr DE=BE rArr DA=DE=BE.`
Now, `BC=BE+EC=AD+AC.`
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