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Question

In ABC,ACB=90o Seg CD side AB and seg DE side CB show that CD2×AC=AD×AB×DE
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Solution

We have,

ABC is a right angle triangle.

In which ACB=90o

And CDABandDEAB

Where D is any point in AB and E is any point in CB.

Prove that:- CD2×AC=AD×AB×DE

Proof:-

InΔACBandΔADC

CABDAC(Reflexive)

ACBADC(Rightangle)

Thus, By AA similarity.

Then,

ΔACBΔADC......(1)

ACAD=ABAC

AC2=AB×AD......(2)

Now, Similarly

ΔACBΔCDB......(3)

Now,

InΔCEDandΔCDB

ECDDCB(Reflexive)

CEDCDB(Rightangle)


By AA similarity.

ΔCEDΔCDB.......(4)

From equation (3) and (4) to,

ΔACBΔCED......(5)

By equation (1) and (5) to, we get,

ΔADCΔCED

ACCD=DCED

CD2=AC×DE......(6)

From equation (2) to,

AC2=AB×AD

On multiplying both side by CD2 and we get,

CD2×AC2=CD2×AB×AD

CD2×AC2=AC×DE×AB×AD

CD2×AC=DE×AB×AD

CD2×AC=AD×AB×DE

Hence proved.


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