# Triangle ABC is an isosceles triangle with AB=AC, side BA is produced to D such that AB=AD. Prove that angle BCD is a right angle.

In triangle ABC

AB=AC  so,angle ABC=angleACB

Now let angleABC=angleACB=x and angleACD=angleADC=y

so, ABC+BCD+BDC=180

>x+x+y+y=180

>2[x+y]=180

>x+y=90

>BCD=90

Therefore angle BCD is a right angle

• 78
M A T H S = Meri Atma Tughe Hamesha Sataygi
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This is the solution for question if AB||CD.prove that p+q-r=180° • -2
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angle bcd = angle acb + angle acd
Now, angle acb = angle abc and angle acd = angle adc (Isosceles triangle)
angle bcd = angle abc + angle adc
angle bcd = 180 - angle bcd          ( Angle sum property)
angle bcd * 2 = 180
angle bcd = 90
Hence proved.

Cheers*
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hi
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In the figure angle AOC and angleBOC form a linear pair . If a-2b=30degree find 'a' and 'b'
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Answer : Given: AB = AC and AD = AB To Prove: ∠ BCD is a right angle Proof: In AB = AC (Given) ∠ACB = ∠ABC (Angles opposite to equal sides are equal) In AD = AB (Given) ∠ADC = ∠ACD (Angles opposite to equal sides are equal) Now, In ∠CAB + ∠ACB + ∠ABC = 180o ( Sum of interior angles of a triangle) ∠CAB + 2 ∠ACB = 180o (∠ACB = ∠ABC) ∠CAB = 180o - 2 ∠ACB (i) Similarly, In ∠CAD = 180o - 2 ∠ACD (ii) Also, ∠CAB + ∠CAD = 180o (BD is a straight line) Adding (i) and (ii), we get ∠CAB + ∠CAD = 180o - 2∠ACB + 180o - 2∠ACD 180o = 360o - 2∠ACB - 2∠ACD 2 (∠ACB + ∠ACD) = 180o ∠BCD = 90o Hence, Proved
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