Question

question 27 please its urgent

Answer 1

I have no feasibility of making a sketch and provide the explanation; so kindly mak your own sketch and follow the below mentioned steps. Given: i) ABC is a triangle, in which AC > AB; this is only for the inference that the altitude from A on to BC lies towards B, that is point 'E' lies between B and D; this is the most important inference for proving the given result. ii) D is the midpoint of BC iii) AE perpendicular BC Required to prove: AB^2=AD^2-BC*DE+1/4BC^2 PROOF: [for easy convenience, let me denote all squares by * and multiplication by 'x'] 1) Since AE is perpendicular to BC, ABE is a right triangle with angle E = 90 deg. 2) Hence applying Pythagoras theorem we have, AB* = AE* + BE* 3) In similar way from the right triangle, ADE, AD* = AE* + DE*, => AE* = AD* - DE* 4) Substituting for AE* from step 3 in step 2, we have, AB* = AD* - DE* + BE* 5) From the figure (Kindly make according to the description given in data), BE = BD - DE 6) Substituting this in (4), AB* = AD* - DE* + (BD-DE)* 7) Expanding, AB* = AD* - DE* + BD* + DE* - 2 x BD x DE 8) Cancelling -DE* and +DE* and substituting BD = (1/2)BC {Since, D is mid point of BC} we get, AB* = AD* + (1/4)BC* - BC x DE Thus it is proved that "AB^2=AD^2-BC*DE+(1/4)BC^2"

Answer 2

Aliya thanks for answering this one even I was looking for this question