Q5 of 26 Page 86

In the figure 6.20, P is the center of the circle. chord AB and chord CD intersect on the diameter at the point E If ∠AEP ≅ ∠DEP then prove that AB = CD.



In the figure, we join PA and PD. Draw perpendiculars on AB and CD from P as PM and PN respectively.


∠AEP = ∠ DEP (given)


So, ∠PEN = ∠PEM (M and N are points on line AE and ED respectively)


In ΔPEN and ΔPEM,


∠PNE = ∠PME = 90°


∠PEN = ∠PEM


PE = PE (common)


Therefore, Δ PEN ≅ ΔPEM (by AAS congruence)


∴ PN = PM (by CPCT)


We know that The chords of a circle equidistant from the center of a circle are congruent.


So, AB = CD.


Hence proved.


More from this chapter

All 26 →
2

Construct incircle and circumcircle of an equilateral ΔDSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.

 3

Construct ΔNTS where NT = 5.7 cm, TS = 7.5 cm and ∠NTS = 1100 and draw incircle and circumcircle of it.

 4

In the figure 6.19, C is the center of the circle. seg QT is a diameter CT = 13, CP = 5, find the length of chord RS.

 6

In the figure 6.21, CD is a diameter of the circle with center O. Diameter CD is perpendicular to chord AB at point E. Show that ΔABC is an isosceles triangle.