Advertisements
Advertisements
प्रश्न
In the figure, ∠CPD = ∠BPD and AD is the bisector of ∠BAC. Prove that ΔCAP ≅ ΔBAP and CP = BP.
Advertisements
उत्तर
In ΔBAP and ΔCAP
∠BAP = ∠CAP ...(AD is the bisector of ∠BAC)
AP = AP
∠BPD + ∠BPA = ∠CPA + ∠CPA = 180°
∠BPD = ∠CPD
⇒ ∠BPA - ∠CPA
Therefore,
ΔCAP ≅ ΔBAP ...(ASA criteria)
Hence, CP = BP.
APPEARS IN
संबंधित प्रश्न
If ΔDEF ≅ ΔBCA, write the part(s) of ΔBCA that correspond to `bar(DF)`
Prove that the sum of three altitudes of a triangle is less than the sum of its sides.
Mark the correct alternative in each of the following:
If ABC ≅ ΔLKM, then side of ΔLKM equal to side AC of ΔABC is
ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD =
In the pair of triangles in the following figure, parts bearing identical marks are congruent. State the test and the correspondence of vertices by the triangle in pairs is congruent.
In the following figure, OA = OC and AB = BC.
Prove that:
(i) ∠AOB = 90o
(ii) ΔAOD ≅ ΔCOD
(iii) AD = CD
State, whether the pairs of triangles given in the following figures are congruent or not:
In a triangle ABC, if D is midpoint of BC; AD is produced upto E such as DE = AD, then prove that:
a. DABD andDECD are congruent.
b. AB = EC
c. AB is parallel to EC
AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.
If ∆PQR is congruent to ∆STU (see figure), then what is the length of TU?
