Advertisements
Advertisements
प्रश्न
O is any point in the ΔABC such that the perpendicular drawn from O on AB and AC are equal. Prove that OA is the bisector of ∠BAC.
Advertisements
उत्तर
In ΔPOA and ΔQOA
∠OPA = ∠OQA = 90°
OP = OQ ...(given)
AO = AO
Therefore, ΔPOA ≅ ΔQOA ...(SSA criteria)
Hence, ∠PAO = ∠QAO
Thus, OA bisects ∠BAC.
APPEARS IN
संबंधित प्रश्न
Complete the congruence statement:
ΔBCA ≅?
ΔQRS ≅?
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ΔADE ≅ΔBCE.
In the pair of triangles given below, the parts shown by identical marks are congruent. State the test and the one-to-one correspondence of vertices by which the triangles in the pair are congruent, the remaining congruent parts.
In the following figure, OA = OC and AB = BC.
Prove that: AD = CD
Prove that:
(i) ∆ ABC ≅ ∆ ADC
(ii) ∠B = ∠D
(iii) AC bisects angle DCB
A is any point in the angle PQR such that the perpendiculars drawn from A on PQ and QR are equal. Prove that ∠AQP = ∠AQR.
In the figure, ∠CPD = ∠BPD and AD is the bisector of ∠BAC. Prove that ΔCAP ≅ ΔBAP and CP = BP.
In the given figure, AB = DB and AC = DC. Find the values of x and y.
“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.” Is the statement true? Why?
Without drawing the triangles write all six pairs of equal measures in the following pairs of congruent triangles.
∆ABC ≅ ∆LMN
