In the adjacent figure, AB is a chord of circle with centre O. CD is the diameter perpendicualr to AB. Show that AD=BD.
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Hi ,
Given : AB is a chord of a circle with center
O . CD is the diameter Perpendicular to AB.
RTP : AD = BD
proof :
Let P is the intersecting point of AB and CD.
From triangles APD and BPD
<APD = <BPD = 90° ( GIVEN )
DP = DP ( Common side )
AP = PB ( diameter bisects chord )
Therefore ,
triangle APD congruent to triangle BPD
( SAS creiteria )
AD = BD ( corrospoding parts of congruent
triangles )
Hence proved.
I hope this helps you.
: )
Given : AB is a chord of a circle with center
O . CD is the diameter Perpendicular to AB.
RTP : AD = BD
proof :
Let P is the intersecting point of AB and CD.
From triangles APD and BPD
<APD = <BPD = 90° ( GIVEN )
DP = DP ( Common side )
AP = PB ( diameter bisects chord )
Therefore ,
triangle APD congruent to triangle BPD
( SAS creiteria )
AD = BD ( corrospoding parts of congruent
triangles )
Hence proved.
I hope this helps you.
: )
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