Question

prove that the two circles drawn on the two equal sides of an isosceles triangle as diameters pass through the mid point of the third side​

Answer 1

Answer:

Step-by-step explanation:

Given:

Two circles are drawn on the two equal sides of an isosceles triangle ΔABC.

The two circles intersect each other at A and D.

∴ AB = AC

Construction: Join AD.

To Prove : BD = CD

Proof:

In ΔADB and ΔADC

∠1 = ∠2 (Angle opposite to equal sides are equal)

∠ADB = ∠CDA = 90° (Angle in a semi-circle is 90°)

∴ ΔADB ~ ΔADC (AA Similarity)

⇒ $\frac{AB}{AC} = \frac{AD}{AD} = \frac{BD}{DC}$ (If two triangles are similar, then their sides are proportional)

⇒ 1 =  $\frac{BD}{DC}$    [ $\frac{AD}{AD} = 1 and \frac{AB}{AC} = 1$ (∵ AB = AC(given)]

⇒ BD = DC

Hence proved.

Answer 2

Answer:

Given: △ABC is an isosceles triangle with AB = AC. A circle is drawn taking AB as diameter which intersects the side BC at D

To prove: BD = DC

Construction: Join AD

Proof: ∠ADB=90

o

[Angle in a semi-circle is right angle]

∠ADB+∠ADC=180

o

⇒∠ADC=90

o

In △ABD and △ACD

AB=AC [Given]

∠ADB=∠ADC

AD=AD [Common]

∴△ABD≅△ACD [RHS congruence criterion]

So,

BD=DC [By CPCT]