Advertisements
Advertisements
प्रश्न
In the following figure, BL = CM.
Prove that AD is a median of triangle ABC.
उत्तर
In ΔDLB and ΔDMC,
BL = CM ...( given )
∠DLB = ∠DMC ...( Both are 90° )
∠BDL = ∠CDM ....( vertically opposite angels )
∴ ΔDLB ≅ ΔDMC ....( AAS congruence criterion )
BD = CD ....( cpct )
Hence, AD is the median of ΔABC.
APPEARS IN
संबंधित प्रश्न
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Which congruence criterion do you use in the following?
Given: EB = DB
AE = BC
∠A = ∠C = 90°
So, ΔABE ≅ ΔCDB
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
The following figure shows a circle with center O.
If OP is perpendicular to AB, prove that AP = BP.
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB is parallel to EC.
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
If AP bisects angle BAC and M is any point on AP, prove that the perpendiculars drawn from M to AB and AC are equal.
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.
In the following figure, ∠A = ∠C and AB = BC.
Prove that ΔABD ≅ ΔCBE.