Advertisements
Advertisements
प्रश्न
Two right-angled triangles ABC and ADC have the same base AC. If BC = DC, prove that AC bisects ∠BCD.
Advertisements
उत्तर

In ΔABC and ΔADC
∠BAC = ∠DAC ...(90°)
BC = DC
AC = AC ...(common)
Therefore, ΔABC ≅ ΔADC ...(SSA criteria)
Hence, ∠BCA = ∠DCA
Thus, AC bisects ∠BCD.
APPEARS IN
संबंधित प्रश्न
If ABC and DEF are two triangles such that ΔABC \[\cong\] ΔFDE and AB = 5cm, ∠B = 40°
In the given figure, if AC is bisector of ∠BAD such that AB = 3 cm and AC = 5 cm, then CD =

In the following example, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same sign. Observe the figure and state the test by which the triangle in each pair are congruent.

By ______ test
Δ ABC ≅ ΔPQR
From the information shown in the figure, state the test assuring the congruence of ΔABC and ΔPQR. Write the remaining congruent parts of the triangles.

Prove that:
(i) ∆ ABC ≅ ∆ ADC
(ii) ∠B = ∠D

In the figure, BC = CE and ∠1 = ∠2. Prove that ΔGCB ≅ ΔDCE.
Prove that in an isosceles triangle the altitude from the vertex will bisect the base.
In ΔABC, AD is a median. The perpendiculars from B and C meet the line AD produced at X and Y. Prove that BX = CY.
Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer.
Without drawing the triangles write all six pairs of equal measures in the following pairs of congruent triangles.
∆STU ≅ ∆DEF
