Question

AD/DB=AE/EC and angleADE=AngleACB. prove that the triangle ABC Is an isosceles triangle ​

Answer 1

Answer:

IN ΔABC AND AED

SO, AD/DB = AE/EC (GIVEN)

==> ∠ADE = ∠ACB (GIVEN)

==> ∠BAC = ∠CAB (COMMON)

SO, ΔABC ≡ ΔAED (BY AAS SIMILARITY CRITERIA)

==> ΔABC/ΔAED ≡ 1

SO, ΔABC/ΔAED = 1 ...(1)

GIVEN ==> AD/DB = AE/EC

WE ALSO WRITE THAT

==> AD/AE = DB/EC

SO, WE KNOW THAT AREA OF RATIO OF TWO SIMILAR TRIANGLE IS EQUAL TO THEIR SQUARE OF CORRESPONDING SIDES

===> AR(ABC)/AR(AED) = (AD/AE)² = (BD/EC)²

==> 1 = AD²/AE² = BD²/EC² (PUTTING EQ(1) )

==> SO, AE ²= AD² AND EC² = BD²

===> AE = AD AND EC = BD

SO, ADDING BOTH EQUATION

==> AD+BD = AE+EC

==> AB = AC

SO, TWO SIDES ARE EQUAL

SO, ΔABC IS AN ISOSCELES TRIANGLE