Answer to Question #156530 in Geometry for Mahama

Question

ABCD is a square and P, Q are the midpoints of BC, CD respectively. If AP = a and AQ = b, find in terms of a and b, the directed line segments (i) AB, (ii) AD, (iii) BD and (iv) AC.

Answer 1

To find AB and AD we have to make some equations. First let us sign BP = x, AD = 2x, QD = y, AB = 2y. From the triangle ABP and from triangle AQD we can write two equations using the Pythagoras Theorem:  

$4y^2+x^2=a^2$

$4x^2+y^2=b^2$

$x^2=a^2−4y^2$

$4(a^2−4y^2)+y^2=b^2$

$4a^2−16y^2+y^2=b^2$

$4a^2−15y^2=b^2$

$y^2=(4a^2−b^2)/15$

$y=\sqrt{(4a^2−b^2)/15} ; 2y=2\sqrt{(4a^2−b^2)/15}$

$x^2=a^2−4(4a^2−b^2)/15$

$x^2=(15a^2−16a^2+4b^2)/15$

$x^2=(4b^2−a^2)/15$

$x=\sqrt{(4b^2−a^2)/15} ; 2x=2\sqrt{(4b^2−a^2)/15}$

So $AB=2y=2\sqrt{(4a^2−b^2)/15}$

$AD=2x=2\sqrt{(4b^2−a^2)/15}$

From the triangle ABD we can find BD using the Pythagoras Theorem: 

$BD=AC=2\sqrt{(4a^2−b^2)/15+(4b^2−a^2)/15}=$

$=2\sqrt{(4a^2−b^2+4b^2−a^2)/15}=2\sqrt{(3a^2+3b^2)/15}=2\sqrt{(a^2+b^2)/5}$