Advertisements
Advertisements
प्रश्न
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Advertisements
उत्तर
Given that
∠BAD = ∠EAC
On adding ∠DAC on both sides, we get
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD …(I)
Now, in △ABC and △AED,
AB = AD ...[Given]
AC = AE ...[Given]
∠BAC = ∠EAD ...[By (I)]
∴ △ABC ≌ △ADE ...[By AAS congruence rule]
⇒ BC = DE ...[Corresponding parts of congruent triangles]
APPEARS IN
संबंधित प्रश्न
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
Which congruence criterion do you use in the following?
Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
You want to show that ΔART ≅ ΔPEN,
If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
1) RT = and
2) PN =
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
Explain, why ΔABC ≅ ΔFED.
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔQRP, AB = QR, ∠B = ∠R and ∠C = P.
In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
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.
In the following figure, ∠A = ∠C and AB = BC.
Prove that ΔABD ≅ ΔCBE. 
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
