Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter, bisects the third side of the triangle.
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Let ABC is the isosceles triangle where AB = AC
The circle drawn on AC as diameter intersects BC at D
From Circle theorem we know that the angle inscribed on semicircle is 90 degree
Hence angle ADC is 90 degree
So we have angle ADB is 90 degree
This gives Δ ADC and Δ ADB are right angled triangle.
We have hypotenuse =AB=AC and AD common leg
Two right triangles are congruent if the hypotenuse and one corresponding leg are equal in both triangles.
as.. Δ ADC ΔADB are congruent
∴ BD=DC
The circle drawn on AC as diameter intersects BC at D
From Circle theorem we know that the angle inscribed on semicircle is 90 degree
Hence angle ADC is 90 degree
So we have angle ADB is 90 degree
This gives Δ ADC and Δ ADB are right angled triangle.
We have hypotenuse =AB=AC and AD common leg
Two right triangles are congruent if the hypotenuse and one corresponding leg are equal in both triangles.
as.. Δ ADC ΔADB are congruent
∴ BD=DC
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