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# ABC is a triangle with angle C =90*.CD is drawn &D lies on AB such that angleCDA=angleCDB=90*.?

Prove that AC^2/BC^2=AD/BD

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no proper reply yet. where are the top contributers?

### 4 Answers

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- Sridhar RLv 61 decade agoFavourite answer
I think you have asked this question twice i have already answered this anyhow here is the proof

in triangle ACD

angle ACD + angle CAD = 90

in triangle ACB

angle CAB + ABC = 90

angle CAB = angle CAD same angle

therefore angle ACD = angle ABC --------------------------------(1)

consider the Triangles

ACD and CBD

angle ACD = angle CBD | angle ABC = angle CBD

angle ADC = angle CDB

so the two triangles are similar

so we have

AD/CD = AC/CB = DC/DB -----------------------------------2)

AD/DC = AC/BC

DC/BD = AC/BC | from 2

multiplying both

AD/DC * DC/BD = AC/BC * AC/BC

AD /BD = AC^2/BC^2 prooved

- 1 decade ago
we know that in triangle ABC AC^2/BC^2=DC^2+AD^2/CD^2+BD^2

=>YOUR ANSWER

Source(s): general - Anonymous1 decade ago
Math is dull lol

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