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<HAB=<HAD if ABCD #, I incenter of BCD, H orthocenter of IBD

Source: China Northern MO 2014 p5 CNMO

October 29, 2022
geometryequal anglesparallelogram

Problem Statement

As shown in the figure, in the parallelogram ABCDABCD, II is the incenter of △BCD\vartriangle BCD, and HH is the orthocenter of △IBD\vartriangle IBD. Prove that ∠HAB=∠HAD\angle HAB=\angle HAD. https://cdn.artofproblemsolving.com/attachments/4/3/5fa16c208ef3940443854756ae7bdb9c4272ed.png
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