Question

In the following, Triangle FEC congruent Triangle GBD and angle 1= angle 2. prove that Triangle ADE similar triangle ABC.

Answer 1

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⬇️HERE IS YOUR ANSWER⬇️

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Given ∆ FEC ≈ ∆GBD

SO From CPCT ,WE GET BD= CE

ALSO GIVEN:angle1=angle2 ...(1)

SO FROM BASE ANGLE THEOREM IN ∆ADE WE GET

AD =AE ......(2)

From equation 1 and 2 we get ADBD= AECE, So from converse of B.P.T. we get,

DE ll BC

THEN angle1= angle3 (CORRESPONDING ANGLES AS DE ll BC AND AB IS TRANSVERSAL LINE).

AND

angle2=angle4

(CORRESPONDING ANGLES AS DE ll BC AND AC IS TRANSVERSAL LINE ).

FROM ABOVE TWO EQUATIONS WE CAN SAY THAT ∆ADE ~ ∆ABC (Hence proved)

(ByAArule)

HOPE THIS ANSWER HELPS YOU

MARK AS BRAINLIEST ✌✌

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