In quadrilateral ACBD, AC = AD and AB bisects \(\angle A\) (see Fig. 7.16). Show that \(\triangle ABC \cong \triangle ABD\). What can you say about BC and BD?
They are equal.
ABCD is a quadrilateral in which AD = BC and \(\angle DAB = \angle CBA\) (see Fig. 7.17). Prove that:
(i) \(\triangle ABD \cong \triangle BAC\)
(ii) BD = AC
(iii) \(\angle ABD = \angle BAC\)
(i) \(\triangle ABD \cong \triangle BAC\)
(ii) BD = AC
(iii) \(\angle ABD = \angle BAC\)
AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
CD bisects AB, i.e. D is the midpoint of AB.
Lines \(l\) and \(m\) are two parallel lines intersected by another pair of parallel lines \(p\) and \(q\) (see Fig. 7.19). Show that \(\triangle ABC \cong \triangle CDA\).
\(\triangle ABC \cong \triangle CDA\)
Line \(l\) is the bisector of an angle \(\angle A\) and B is any point on \(l\). BP and BQ are perpendiculars from B to the arms of \(\angle A\) (see Fig. 7.20). Show that:
(i) \(\triangle APB \cong \triangle AQB\)
(ii) BP = BQ or B is equidistant from the arms of \(\angle A\).
(i) \(\triangle APB \cong \triangle AQB\)
(ii) BP = BQ (B is equidistant from the arms of \(\angle A\)).
In Fig. 7.21, AC = AE, AB = AD and \(\angle BAD = \angle EAC\). Show that BC = DE.
\(\angle BAC = \angle DAE\)
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \(\angle BAD = \angle ABE\) and \(\angle EPA = \angle DPB\) (see Fig. 7.22). Show that:
(i) \(\triangle DAP \cong \triangle EBP\)
(ii) AD = BE.
(i) \(\triangle DAP \cong \triangle EBP\)
(ii) AD = BE
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. D is joined to B (see Fig. 7.23). Show that:
(i) \(\triangle AMC \cong \triangle BMD\)
(ii) \(\angle DBC\) is a right angle.
(iii) \(\triangle DBC \cong \triangle ACB\)
(iv) \(CM = \tfrac{1}{2} AB\).
(i) \(\triangle AMC \cong \triangle BMD\)
(ii) \(\angle DBC\) is a right angle.
(iii) \(\triangle DBC \cong \triangle ACB\)
(iv) \(CM = \tfrac{1}{2} AB\)
In an isosceles triangle \(ABC\), with \(AB = AC\), the bisectors of \(\angle B\) and \(\angle C\) intersect each other at \(O\). Join \(A\) to \(O\). Show that: (i) \(OB = OC\), (ii) \(AO\) bisects \(\angle A\).
In \( \triangle ABC \), \(AD\) is the perpendicular bisector of \(BC\). Show that \( \triangle ABC \) is an isosceles triangle in which \(AB = AC\).
\(ABC\) is an isosceles triangle in which altitudes \(BE\) and \(CF\) are drawn to equal sides \(AC\) and \(AB\) respectively. Show that these altitudes are equal.
\(ABC\) is a triangle in which altitudes \(BE\) and \(CF\) to sides \(AC\) and \(AB\) are equal. Show that (i) \( \triangle ABE \cong \triangle ACF \), (ii) \(AB = AC\), i.e., \(ABC\) is an isosceles triangle.
\(ABC\) and \(DBC\) are two isosceles triangles on the same base \(BC\). Show that \(\angle ABD = \angle ACD\).
\(\triangle ABC\) is an isosceles triangle in which \(AB = AC\). Side \(BA\) is produced to \(D\) such that \(AD = AB\). Show that \(\angle BCD\) is a right angle.
\(\angle BCD = \angle BCA + \angle DCA = \angle B + \angle D\)
\(ABC\) is a right-angled triangle in which \(\angle A = 90^\circ\) and \(AB = AC\). Find \(\angle B\) and \(\angle C\).
each is of 45°
Show that the angles of an equilateral triangle are \(60^\circ\) each.
\(\triangle ABC\) and \(\triangle DBC\) are two isosceles triangles on the same base \(BC\) and vertices \(A\) and \(D\) are on the same side of \(BC\). If \(AD\) is extended to intersect \(BC\) at \(P\), show that (i) \(\triangle ABD \cong \triangle ACD\), (ii) \(\triangle ABP \cong \triangle ACP\), (iii) \(AP\) bisects \(\angle A\) as well as \(\angle D\), (iv) \(AP\) is the perpendicular bisector of \(BC\).
\(AD\) is an altitude of an isosceles triangle \(ABC\) in which \(AB = AC\). Show that (i) \(AD\) bisects \(BC\), (ii) \(AD\) bisects \(\angle A\).
Two sides \(AB\) and \(BC\) and median \(AM\) of one triangle \(ABC\) are respectively equal to sides \(PQ\) and \(QR\) and median \(PN\) of \(\triangle PQR\). Show that (i) \(\triangle ABM \cong \triangle PQN\), (ii) \(\triangle ABC \cong \triangle PQR\).
From (i), \(\angle ABM = \angle PQN\)
\(BE\) and \(CF\) are two equal altitudes of a triangle \(ABC\). Using RHS congruence rule, prove that the triangle \(ABC\) is isosceles.
\(ABC\) is an isosceles triangle with \(AB = AC\). Draw \(AP \perp BC\) to show that \(\angle B = \angle C\).