Question

24..If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side​

Answer 1

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If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

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Solution

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Given,

Two circles are drawn on the sides AB and AC of the triangle

ABC as diameters. The circles intersected at D.

Construction: AD is joined.

To prove: D lies on BC. We have to prove that BDC is a straight line.

Proof:

∠ADB=∠ADC=90° ...Angle in the semi circle

Now,

∠ADB+∠ADC=180°

⇒∠BDC is straight line.

Answer 2

Step-by-step explanation:

Steps of construction:

(1) Draw seg BC of length 6cm.

(2) Draw ray BE such that ∠CBE=100

∘

(3) Take point D on the opposite of ray BE such that BD=2.5cm.

(4) Construct the perpendicular bisector of seg DC.

(5) Name the point of intersection of ray BE and the perpendicular bisector of DC as A.

(6) Draw seg AC.

Therefore, ΔABC is required triangle.