Hello math bugs(š) and hivers(š)
I hope you are strong amd stout and doing great in life
Well come to another intersting gemetric problem and its solution. āABC is a random triangle. AD is perpendicular to BC and E is a random point on AD. If *AB =10 cm , AC= 17 cm and CE =16 cm, What is BE? Check the figure down below:
This is a true use of pythagorean triplet .Today we need no other concept except it. If we use pythagorean trip on triangles on the problem figure, a sweet relation must come out and whcih is given by ABĀ²+ CEĀ² = ACĀ² + BEĀ². I'll proof it, once we get to the answer.
Check the following window for value of be BE
Let's prove it
We gonna use pythagorean triplet in āABD and in āCDE.
From āABD
We have ABĀ² = ADĀ² + BDĀ²...equation (1)
From āCDE
We have CEĀ² = CDĀ² + EDĀ²...equation (2)
Now, after adding equations (1) and (2)
We have ABĀ²+ CEĀ² = ADĀ² + CDĀ² + BDĀ² + EDĀ²...eqn (3)
Again, using pythagorean teiplet on āADC and āBDE,
We can say
ADĀ² + CDĀ² = ACĀ² and BDĀ² + EDĀ² = BEĀ²
Now, if we replace ADĀ² + CDĀ² and BDĀ² + EDĀ² in equation (3) by ACĀ² and BEĀ² respectively, we can have
ABĀ² + CEĀ² = ACĀ² + BEĀ² (proved)
Check the following figure also if you missing out something
For replacing check this one
All the figure are used here made by me using android application. There may be some construction misatke(s) or they may not be proper drawing. It takes time to make those figure. If you find any dispute, please ingnore it and cosider the data only.
I hope you find this post interesting and understandable.
Thank you so much for visiting
Have a nice day
All is well
Regards: @meta007