Advertisements
Advertisements
Question
In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA = YC.
Advertisements
Solution
ABCD is a parallelogram in which ∠A and ∠C are obtuse.
Points X and Y are taken on the diagonal BD.
Such that ∠XAD = ∠YCB = 90°.
We need to prove that XA = YC
Proof:
ln ΔXAD and ΔYCB
∠XAD = ∠YCB= 90° ...[ Given ]
AD = BC ...[ Opposite sides of a parallelogram ]
∠ADX = ∠CBY ...[ Alternate angles ]
∴ By Angle-Side-Angle criterion of congruence,
ΔXAD ≅ ΔYCB
The corresponding parts of the congruent triangles are congruent.
∴ XA = YC ...[ c.p.c.t. ]
Hence proved.
APPEARS IN
RELATED QUESTIONS
Which congruence criterion do you use in the following?
Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
Which congruence criterion do you use in the following?
Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
In Fig. 10.92, it is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C =∠F. Are the two triangles necessarily congruent?
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5cm and ∠D = 75°. Are two triangles congruent?
D, E, F are the mid-point of the sides BC, CA and AB respectively of ΔABC. Then ΔDEF is congruent to triangle
The given figure shows a circle with center O. P is mid-point of chord AB.
Show that OP is perpendicular to AB.
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
