Q. 64.0( 25 Votes )

# 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.

Answer :

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

So, AE = EB

In ΔACE and ΔBCE,

AE = EB

∠AEC = ∠BEC = 90°

CE = CE (common)

Δ ACE ≅ Δ BCE (By SAS congruence)

Therefore, AC = BC (by CPCT)

Hence proved that ABC is an isosceles triangle.

Rate this question :

In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the center. Find the lengths of the chords.

Seg PM and seg PN are congruent chords of a circle with center C. Show that the ray PC is the bisector of ∠NPM.

MHB - Math Part-IIIn 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.

MHB - Math Part-II

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.

MHB - Math Part-II

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.

MHB - Math Part-II