সমান্তরাল ABCD-এ, সন্নিহিত কোণের দ্বিখণ্ডক A এবং D পরস্পরকে P বিন্দুতে ছেদ করে।

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that D P = B Q . Show that: (i) DeltaA P D~= DeltaC Q B (ii) A P = C Q (iii) DeltaA B C (iv) A Q = C P (v) APCQ is a parallelogram.

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that a r" "(A P B)" "xx" "a r" "(C P D)" "=" "a r" "(A P D)" "xx" "a r" "(B P C)dot

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P . Prove that ar(triangleABP)=ar("qaud. ABCD") .

In the figure, ABCD is a quadrilateral. A line DP drawn parallel to diagonal AC from point D, meet BC produced at P. Prove that: are of Delta ABP = " area of " square ABCD

Consider a rectangle ABCD formed by the points A=(0,0), B= (6, 0), C =(6, 4) and D =(0,4), P (x, y) is a moving interior point of the rectangle, moving in such a way that d (P, AB)le min {d (P, BC), d (P, CD) and d (P, AD)} here d (P, AB), d (P, BC), d (P, CD) and d (P, AD) represent the distance of the point P from the sides AB, BC, CD and AD respectively. Area of the region representing all possible positions of the point P is equal to

In Figure,bisectors of /_B and /_D of quadrilateral ABCD meet CD and AB produced at Pand Q respectively.Prove that : /_P+/_Q=(1)/(2)(/_ABC+/_ADC)

Let ABCD be a square. An arc of a circle with A as centre and AB as radius is drawn inside the square joining the points B and D. Points P on AB, S on AD, Q and R on arc BD are taken such that PQRS is a square. Further suppose that PQ and RS are parallel toAC. Then, (area PQRS)/(area ABCD) is