Question

abc is triangle in which ab=ac and d is the point on acsuch that bc2 =acXcd . prove that BD +BC

Answer 1

Answer:

In △ABC AB=AC

⇒∠B=∠C (Angles opposite to equal sides are equal)

Now using angle sum property

∠A+∠B+∠C=180

∘

⇒80

∘

+∠C+∠C=180

∘

⇒2∠C=180

∘

−80

∘

⇒∠C=

2

100

∘

=50

∘

now ∠C+∠x=180

∘

(Angles made on straight line (AC) are supplementary)

⇒50

∘

+∠x=180

∘

⇒∠x=180

∘

−50

∘

=130

∘

Answer 2

Answer:

given

in triangle abc

a b equal to AC and D is a point on AC

such that.

BC × BC= AC×AD

we are to prove BC equal to DC

proof

BC×BC=AC×AD

BC/CD=AC/BC=∆ABC

Simalar∆BDC

angle BAC equal to Angle DBC ( corresponding angle)

angle abc equal to angle BCD (corresponding angle)

angle ACD equal to angle DCB (corresponding angle)

again in triangle abc

AB equal to AC

angle abc equal to angle ACB equal to angle DCB

therefore triangle bdc angle bdc equal to BCD

therefore BD equal to BC