ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that
(i) `DeltaAPB ~= DeltaCQD`
(ii) `AP = CQ`

ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD. Show that DeltaAPB approx DeltaCQD AP=CQ

ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal (see figure) . Show that (i) DeltaABD ~= DeltaACE (ii) AB = AC i.e., ABC is an isosceles triangle.

In right triangle ABC, right angle is 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. Point D is joined to point B (see figure). Show that : (i) DeltaAMC ~= DeltaBMD (ii) /_DBC is a right angle (iii) Delta DBC ~= DeltaACB (iv) CM = 1/2 AB .

ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are the points of tri section of the diagonal BD. Prove that CQ" ||" AP and also AC bisects PQ (see figure).

In a parallelogram ABCD, E and F are the mid points of sides AB and CD respectively (see fig.) Show that the line segements AF and EC trisect the diagonal BD.

P is a point equidistant from two lines l and m intersecting at point A (see figure). Show that the line AP bisects the angle between them.

ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD "||" BA as shown in the figure. Show that (i) angleDAC = angleBCA (ii) ABCD is a parallelogram

In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect AD = CQ. If AQ intersect DC at P, Show that ar (BPC) = ar (DPQ)

triangleABC and triangleDBC 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) triangleABD ~= triangleACD (ii) triangleABP ~= triangleACP (iii) AP bisects angleA as well as angleD (iv) AP is the perpendicular bisector of BC.

In the given figure, diagonals AC and BD of quadrilateral ABCD interset at O such that OB = OD. If AB = CD, then show that : (i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB) (iii) DA || CB or ABCD is a parallelogram .